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Boundary value problem with substitution

  1. Sep 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the general solution to the boundary value problem.

    2. Relevant equations
    [tex]
    (xy')' + \lambda x^{-1}y = 0
    [/tex]
    [tex]y(1) = 0[/tex]
    [tex]y(e) = 0[/tex]
    use [tex]x = e^t[/tex]

    3. The attempt at a solution

    [tex]x = e^t[/tex] so [tex]\frac{dx}{dt} = e^t[/tex]

    using chain rule:
    [tex]y' = e^{-t}\frac{dy}{dt}[/tex]

    Substituting this in:

    [tex]\frac{d}{dx}(e^t(e^{-t}\frac{dy}{dt})) + \lambda e^{-t}y = 0[/tex]
    [tex]\frac{d}{dx}(\frac{dy}{dt}) = \lambda e^{-t}y = 0[/tex]
    [tex]\frac{d}{dx}(y'e^t) + \lambda e^{-t}y = 0[/tex]

    From this point, I feel like I am going in circles. I want to get everything in equal powers of [tex]e[/tex], so I can cancel it out, get a general solution in terms of [tex]t[/tex], and plug in the boundary values to find my constants.
     
  2. jcsd
  3. Sep 12, 2010 #2

    hunt_mat

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    Expand out the ODE fully:
    [tex]
    \frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}+\frac{\lambda}{x}y\Rightarrow x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+\lambda y=0
    [/tex]
    This is called the Euler equation, look for solutions of the form x^{n}.
     
  4. Sep 12, 2010 #3
    How did you arrive at that?

    Here is my attempt at expanding it out fully:

    [tex]\frac{d}{dx}(x\frac{dy}{dx}) + \frac{\lambda}{x}y = 0[/tex]
    [tex]\frac{dy}{dx} + x\frac{d^2y}{dx^2} + \frac{\lambda}{x}y = 0[/tex]
    [tex]x^2y'' + y'x + \lambda y = 0[/tex]

    ... Ah, I see what you did, ok. Thanks for the tip. Do I even need a substitution with this?
     
  5. Sep 12, 2010 #4
    I continued the problem, using the method you suggested, and I ended up with the following differential equation, in terms of t:

    [tex]\ddot{y} + \lambda y = 0[/tex]

    The general solution to this, in terms of [tex]t[/tex], is:

    [tex]y = A\cos{\sqrt{\lambda}t} + B\sin{\sqrt{\lambda}t}[/tex]

    Boundary conditions holding true, I get:

    [tex]A = 0[/tex]
    [tex]B[/tex] is arbitrary

    [tex]y(x) = Bsin(n\ln{x})[/tex] where [tex]n = 1, 2, 3, \ldots[/tex]

    If you want, I can be more thorough as to how I got this answer, especially if it's wrong!
     
  6. Sep 12, 2010 #5

    hunt_mat

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    As I said sibstitute in y=x^{n} to obtain the following:
    [tex]
    \left[ n(n-1)+n+\lambda\right] x^{n}=0
    [/tex]
    We require an equation for n, just like the constant coefficient case, can you say what this equation is?

    I think that in general you're over thinking it. You don't need a substitution for this equation.
     
  7. Sep 12, 2010 #6
    Using the substitution method, when solving using the second boundary condition, I am ending up with a discrepancy when attempting a solution with both methods, and it's minor.

    For the method of substituting like what the problem suggests, I get:

    [tex]0 = B\sin{\sqrt{\lambda}}[/tex]

    So...

    [tex]\lambda = n^2\pi^2[/tex]

    Using the euler format, I am getting...

    [tex]0 = B\sin{n\sqrt{\lambda}}[/tex]

    So...

    [tex]\lambda = \frac{m^2pi^2}{n^2}[/tex]

    where [tex]m[/tex], in this case, is [tex]n[/tex] in the first method.

    I checked my algebra, and I don't see how these can be so different. Is it because [tex]n[/tex] and [tex]m[/tex] are arbitrary, and can combine to equal a single [tex]n[/tex]?
     
  8. Sep 12, 2010 #7

    hunt_mat

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    The two solutions are:
    [tex]
    n=\pm\sqrt{\lambda}
    [/tex]
    So the solution becomes:
    [tex]
    y=Ax^{\sqrt{\lambda}}+Bx^{-\sqrt{\lambda}}
    [/tex]
    Then just use the boundary conditions to find A and B.
     
  9. Sep 12, 2010 #8
    Really? So, the solution is not periodic at all? That's strange. I kept getting imaginary roots, which, by doing it two different methods, resulted in getting a sine and cosine. I will post my work in a little bit.
     
  10. Sep 13, 2010 #9

    hunt_mat

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    I made a mistake in my algebra, my solutions should have been
    [tex]
    n=\pm i\sqrt{\lambda}
    [/tex]
    Then my solution reduces to yours... The solution my method becomes:
    [tex]
    y=\sum_{n=0}^{\infty}A_{n}\left( x^{n\pi i}-x^{-n\pi i}\right)
    [/tex]
     
  11. Sep 13, 2010 #10
    That's actually an interesting notation. It seems to be more properly descriptive of what all the solutions are.
     
  12. Sep 13, 2010 #11

    hunt_mat

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    By writing
    [tex]
    x^{k}=e^{k\log x}
    [/tex]
    My solutions become
    [tex]
    y=\sum_{n=0}^{\infty}B_{n}\sin (n\pi\log x)
    [/tex]
     
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